Question

find the zeros of the quadratic polynomials 6xsquare -3 - 7x verify relationship between zeros and coefficiants​

Answer 1

Answer:

$\mathtt{SOLUTION:-}$

$\mathtt \purple{ {6x}^{2} - 3 - 7x = 0 }$

$\mathtt \purple{ {6x}^{2} - 7x - 3 = 0 }$

$\mathtt \purple{ {6x}^{2} + 2x - 9x - 3 = 0 }$

$\mathtt \purple{2x( {3x} - 1) - 3(3x - 1) = 0 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt \purple{(2x - 3) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: (3x - 1) = 0}$

$\mathtt \purple{2x = 0 + 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3x = 0 + 1}$

$\mathtt \purple{x = \frac{ 3}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{1}{3} }$

$\mathtt \purple{ \alpha = \frac{ 3}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \beta = \frac{1}{3} }$

$\mathtt{SUM \: OF \: ZEROS}$

$\mathtt \purple{ \alpha + \beta = \frac{ - b}{a} }$

$\mathtt \purple{ \frac{ 3}{2} + \frac{1}{3} = \frac{ -( - 7)}{6} }$

$\mathtt \purple{ \frac{ 9 + 2}{6 } = \frac{ 7}{6} }$

$\mathtt \purple{ \frac{ 7}{6} = \frac{ 7}{6} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\mathtt{PRODUCT \: OF \: ZEROS}$

$\mathtt \purple{ \alpha \beta = \frac{c}{a} }$

$\mathtt \purple{ \frac{ - 3}{2} \times \frac{1}{3} = \frac{ - 3}{6} }$

$\mathtt \purple{ \frac{ - 3}{6} = \frac{ - 3}{6} }$